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Finding the reason behind the value of the integral.
Not sure how to solve this integralTI Nspire CX CAS fails to perfrom basic integrationIndefinite integral of absolute valueHelp with Differentiating through the Integral SignInitial value problem without explicit constant findingDefinite integration using even and odd functionsEvaluating a real definite integral using residue theoremHow to plot the graph of $[|sin x|+|cos x|]$ where [.] is the G.I.F?Definite integral involving polylogarithmWhy can't the indefinite integral $intfracsin(x)xmathrm dx$ be found?
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I was just trying to find $$int_0^pi / 2fracsin9xsinx,dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.
I can't figure out the reason and would like to know why this is happening.
definite-integrals
$endgroup$
add a comment |
$begingroup$
I was just trying to find $$int_0^pi / 2fracsin9xsinx,dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.
I can't figure out the reason and would like to know why this is happening.
definite-integrals
$endgroup$
7
$begingroup$
The following identity seems like it may help:$$fracsin((n+1/2)thetasin(theta/2)=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
5 hours ago
add a comment |
$begingroup$
I was just trying to find $$int_0^pi / 2fracsin9xsinx,dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.
I can't figure out the reason and would like to know why this is happening.
definite-integrals
$endgroup$
I was just trying to find $$int_0^pi / 2fracsin9xsinx,dx $$ using an online integral calculator. And surprisingly I found that if I replace $9x$ by $ x,3x,5x$ which are some odd multiples of $x$ the value of integral came out to be $dfrac pi 2$.
I can't figure out the reason and would like to know why this is happening.
definite-integrals
definite-integrals
edited 4 hours ago
Abcd
3,16431339
3,16431339
asked 5 hours ago
JasmineJasmine
368213
368213
7
$begingroup$
The following identity seems like it may help:$$fracsin((n+1/2)thetasin(theta/2)=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
5 hours ago
add a comment |
7
$begingroup$
The following identity seems like it may help:$$fracsin((n+1/2)thetasin(theta/2)=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
5 hours ago
7
7
$begingroup$
The following identity seems like it may help:$$fracsin((n+1/2)thetasin(theta/2)=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
5 hours ago
$begingroup$
The following identity seems like it may help:$$fracsin((n+1/2)thetasin(theta/2)=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint
Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$
$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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active
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$begingroup$
Hint
Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$
$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.
$endgroup$
add a comment |
$begingroup$
Hint
Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$
$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.
$endgroup$
add a comment |
$begingroup$
Hint
Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$
$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.
$endgroup$
Hint
Consider $I(n)=int_0^pi/2 fracsin(nx)sin x dx$
$$I(2m+1)-I(2m-1)=int_0^pi/2 fracsin(2m+1)x-sin(2m-1)xsinx dx=int_0^pi/2 frac2sin(x)cos(2mx)sinx dx$$
$$implies 2int_0^pi/2 cos(2mx)dx$$
Now think what happens to this integral when $m$ is an integer.
And also try to use the fact $I(1)=fracpi2$.
edited 34 mins ago
answered 5 hours ago
NewBornMATHNewBornMATH
44410
44410
add a comment |
add a comment |
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7
$begingroup$
The following identity seems like it may help:$$fracsin((n+1/2)thetasin(theta/2)=1+2cos x+2cos(2x)+cdots+2cos(nx).$$ (This is known as the Dirichlet kernel, and a proof may be found at the corresponding Wikipedia page here.)
$endgroup$
– Semiclassical
5 hours ago